Basic circuit homework

Thread Starter

Sanctuary10

Joined Oct 4, 2020
7
Hi guys,

I'm aware this question is probably super easy but help would be appreciated. I guess I am just confused by the arrangement and how to go about doing it.

I've added a picture of the circuit.

V1=30v (the lower one incase it's not clear)
V2=10v

Find the amp meter reading?
Find the value of Rx?20201003_203321.jpg

Many thanks for any guidance.
 

WBahn

Joined Mar 31, 2012
26,301
The problem really doesn't make sense. It shows an open switch, yet a voltage across a resistor.

Assuming they meant you to assume that the switch is closed, does it make sense to you that if you connect a 30 V battery to two resistors in series that you are going to get 40 V across one of the resistors and 10 V across the other?
 

Thread Starter

Sanctuary10

Joined Oct 4, 2020
7
The problem really doesn't make sense. It shows an open switch, yet a voltage across a resistor.

Assuming they meant you to assume that the switch is closed, does it make sense to you that if you connect a 30 V battery to two resistors in series that you are going to get 40 V across one of the resistors and 10 V across the other?
Yes assume the switch is closed.

The two voltages and how they operate is where I am confused, no it doesnt make sense but it came from trying to follow the flow direction, I'm pretty much just guessing because I don't really know how to approach it. I haven't been given any guidance by my teacher just got handed it on friday. We have done is series and parallel circuits so I assume this is the same but I want to understand the method for the arrangement.

Is it a case of adding the voltages for the entire circuit than applying ohms law as you would for series?

Tha is for the reply.
 

Thread Starter

Sanctuary10

Joined Oct 4, 2020
7
I=20/5 = 4A
Rx = 20/4 = 5ohms

Came to this by dividing the voltage at the junctions, flowing in an anti clockwise direction.
 

Martin_R

Joined Aug 28, 2019
28
I=20/5 = 4A
Rx = 20/4 = 5ohms

Came to this by dividing the voltage at the junctions, flowing in an anti clockwise direction.
Well, almost there! The 5 ohm resistor drops 20 volts as you say, so from ohm's law I=V/R, so amps = 20/5 = 4 amp.
Now for Rx we know 10 volts are across it with 4 amps flowing through it, so now R = V/I , so 10/4 = 2.5 ohm
 

Thread Starter

Sanctuary10

Joined Oct 4, 2020
7
Well, almost there! The 5 ohm resistor drops 20 volts as you say, so from ohm's law I=V/R, so amps = 20/5 = 4 amp.
Now for Rx we know 10 volts are across it with 4 amps flowing through it, so now R = V/I , so 10/4 = 2.5 ohm
Thanks for the reply! One question on that though, how is it 10v across Rx? Should that not drop also? Maybe I'm over thinking it or I'm just a bit thick.
 

MrChips

Joined Oct 2, 2009
22,097
Semantics.
Stop using the word "drop" in circuits. Resistors do not "drop" voltage.

10V is across Rx. That was given to you.
20V is across the 5Ω resistor because the two voltages must add up to 30V which was also given.
 

MrChips

Joined Oct 2, 2009
22,097
Your response in post #3 appears to be wild guesses.

Instead, write down what you know:

1) 30V = I x (5Ω + Rx)

2) 10V = I x Rx

Write this out as two equations:

1a) 30 = I x ( 5 + R )
1b) 30 = ( I x 5 ) + ( I x R )

2a) 10 = I x R

Solve for I and R.
 

Thread Starter

Sanctuary10

Joined Oct 4, 2020
7
Your response in post #3 appears to be wild guesses.

Instead, write down what you know:

1) 30V = I x (5Ω + Rx)

2) 10V = I x Rx

Write this out as two equations:

1a) 30 = I x ( 5 + R )
1b) 30 = ( I x 5 ) + ( I x R )

2a) 10 = I x R

Solve for I and R.
You're not wrong about that, for some odd reason I saw it as if v1 and v2 were voltage inputs rather than readings ^^. Which is why I thought they would interact in some way. Feel silly now I understand it. Obviously I'm a complete newbie when It comes to electronics! Thanks very much everyone! :)
 

WBahn

Joined Mar 31, 2012
26,301
Yes assume the switch is closed.

The two voltages and how they operate is where I am confused, no it doesnt make sense but it came from trying to follow the flow direction, I'm pretty much just guessing because I don't really know how to approach it. I haven't been given any guidance by my teacher just got handed it on friday. We have done is series and parallel circuits so I assume this is the same but I want to understand the method for the arrangement.

Is it a case of adding the voltages for the entire circuit than applying ohms law as you would for series?

Tha is for the reply.
Voltage and currents have polarity -- ask anyone that has ever tried to jump start a car without taking that into account.

What direction is the current flowing in the circuit? Consider which end of the battery is the positive terminal.

The 10 V across the unknown resistance is just the magnitude of the voltage, it doesn't say which end of the resistor is at a higher voltage than the other, just the magnitude of the difference. From the direction of current flow, which end is more positive than the other?

Although you don't know what the magnitude of the voltage across the 5 Ω resistor is, you do know the direction of the current flowing in it, so which end of that resistor is more positive than the other?

Once you have all of the polarities, you can apply Kirchhoff's Voltage Law around the loop to determine the voltage across the 5 Ω resistor.
 

WBahn

Joined Mar 31, 2012
26,301
Well, almost there! The 5 ohm resistor drops 20 volts as you say, so from ohm's law I=V/R, so amps = 20/5 = 4 amp.
Now for Rx we know 10 volts are across it with 4 amps flowing through it, so now R = V/I , so 10/4 = 2.5 ohm
PLEASE do not just work someone's homework for them. That does not do the poster any good -- in fact it usually does harm because they walk away with a sense that they understand when, in reality, it is usually the case that they do not. They then risk discovering this when they attempt to work a similar problem on the exam and can't.
 

Martin_R

Joined Aug 28, 2019
28
PLEASE do not just work someone's homework for them. That does not do the poster any good -- in fact it usually does harm because they walk away with a sense that they understand when, in reality, it is usually the case that they do not. They then risk discovering this when they attempt to work a similar problem on the exam and can't.
My intention was to go through the problem in small understandable chunks, not just to give the answer and do his homework for him. I believe I went through it so he could comprehend how the answer was obtained. But I will take on board what you're saying.
 

Thread Starter

Sanctuary10

Joined Oct 4, 2020
7
PLEASE do not just work someone's homework for them. That does not do the poster any good -- in fact it usually does harm because they walk away with a sense that they understand when, in reality, it is usually the case that they do not. They then risk discovering this when they attempt to work a similar problem on the exam and can't.
To be fair he wasn't doing my homework for me, he helped along with posts after it to point out my errors and misconception. As you should be able to see.. I was asking for explanation after this post to gain the understanding, I wasn't just looking for the answers to copy. I was confused and that helped me realise where I was going wrong.
 

djsfantasi

Joined Apr 11, 2010
6,962
To be fair he wasn't doing my homework for me, he helped along with posts after it to point out my errors and misconception. As you should be able to see.. I was asking for explanation after this post to gain the understanding, I wasn't just looking for the answers to copy. I was confused and that helped me realise where I was going wrong.
To be fair, a more experienced forum member knows how to provide the same help, without performing any calculations for you by asking leading questions.

That’s the proper way to respond to a homework help question. New members are, well, new and also need guidance in responding to homework help questions. Otherwise, this sub-forum might as well be named “Homework Done for You”.

Respectively commented.
 
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